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Motion Under Gravity

A body is dro...

Question

A body is dropped from the top of a tower It acquires a velocity $20 m s^{- 1}$ on reaching the ground. Calculate the height of the tower (Take $g = 10 m s^{- 2}$ )

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Solution

Given,
$u = 0 m / s$
$v = 20 m / s$
$g = 10 m / s^{2}$
From 3rd equation of motion,
$2 g h = v^{2} - u^{2}$
$h = \frac{v^{2} - u^{2}}{2 g}$
$h = \frac{20 \times 20 - 0 \times 0}{2 \times 10}$
$h = 20 m$
The height of the tower is $20 m$ .

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